For a continuous function going through $(-1,0),(1,0),(2,3)$, show there are two distinct fixed points 
Let $f : \mathbb{R}→\mathbb{R}$ be a continuous function such that
$f(−1) = f(1) = 0, f(2) = 3$.
Show that there are distinct $a,b \in\mathbb{R}$ such that $f(a) = a$ and $f(b) = b$.

I'm pretty sure I use the Intermediate Value Theorem here, and to ensure $a \not = b$, I think I would divide this problem up into two intervals $ [-1,1]$ and $[1,2]$. But I have no idea how to do the proof.
 A: Note that if $a,b$ satisfy $f(a) = a$ and $f(b) = b$, then $a,b,f$ satisfy the property whereby $(a,f(a))$ and $(b,f(b))$ lie on the line $y=x$.
Perhaps a trivial observation, but it may make visualization easier. After noticing this, graphing the points and the line gives us the following:

For the case of the interval $[1,2]$, the solution is fairly trivial. If $f$ is continuous, by the intermediate value theorem, you get some value $a$ whereby $a = f(a)$.
For the case of the interval $[-1,1]$, consider three cases. Obviously one of these must be true:
$$f(0) = 0 \qquad f(0) > 0 \qquad f(0) < 0$$
In each case, you can use the obviousness of it ($f(0) = 0$) or the intermediate value theorem ($f(0) \ne 0$) to get a second point.
For this second one in particular, I added a purple point to the graph at the point $(0,p)$; you can vary $p$ in this Desmos demo to help you visualize the situation better.
A: The question says that the value of $f(x)$ is equal to $0$ at 2 points. So we can say the degree of the equation must be atleast 2. Let's assume the degree of the equation to be $2$.$$f(x)=ax^2+bx+c$$ Putting $x=1$,$$a+b+c=0$$ Putting $x=-1$, $$a-b+c=0$$ Putting $x=2$,$$4a+2b+c=3$$ Solving these equations you will get $a=1,b=0,c=(-1)\Rightarrow f(x)=x^2-1$
Now you need to prove that there exists $a,b$ such that $f(a)=a$ and $f(b)=b$
$\Rightarrow x^2-1=x\Rightarrow x^2-x-1=0 \Rightarrow x=\frac{1±\sqrt{5}}{2}\Rightarrow a=\frac{1+\sqrt{5}}{2}, b=\frac{1-\sqrt{5}}{2}$
I am not very sure about this solution..... If you find any error please comment.
